# Sequences and Series – Exercise 9.3[11-20] – Class XI

11. Evaluate 1. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Solution:

Let a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 —-(1)

(a/r)(a)(ar) = 1 —–(2)

From (2), we obtain a3 = 1

a = 1

Substuting a = 1 in equation (1), we get

1/r + 1 + r = 39/10

1 + r + r2 = 39/10 r

10 + 10r + 10r2 – 39r = 0

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r – 5)-2(2r – 5)= 0

(5r – 2)(2r – 5) = 0

r = 2/5 or 5/2

Thus the three terms of G.P. are 5/2, 1 and 2/5

1. How many terms of G.P. 3, 32,33 … are needed to give the sum 120?

Solution:

The given G.P. is 3, 32, 33,…

Let n terms of this G.P. be required to obtain the sum as 120. Thus, four terms of the given G.P. are required to obtain the sum as 120.

14: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum of n terms of the G.P.

Solution:

Let the G.P. be a, ar, ar2,…

According to the given condition,

a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128

⇒ a (1 + r + r2) = 16 ……………. (1)

ar3(1 + r + r2) = 128 …………….. (2)

Dividing equation (2) by (1), we obtain 1. Given a G.P. with a = 729 and 7th term 64, determine S7

Solution:

a = 729 and a7= 64

Let r be the common ratio of the G.P. It is known that

an = a.rn-1

a7 = a.r7-1 =(729)r6

64 = 729(r6)

r6 = 64/729

r6 =(2/3)6

r = 2/3

Also it is known that, 1. Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.

Solution:

Let a be the first term and r be the common ratio of the G.P.

According to the given conditions, Thus the required G.P. is –4/3, –8/3, –16/3,… or 4, -8, 16, -32,…

1. If the 4th ,10th and 16th terms of a G.P. are x , y and z respectively. Prove that x , y, z are in G.P.

Solution:

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a4 = ar3 = x ———-(1)

a10 = ar9 = y ———-(2)

a16 = ar15 = z ———-(3)

Dividing (2) by (1), we get

y/x = (ar9)/(ar3)

y/x = r6

Dividing (3) by (2), we get

z/y = (ar15)/(ar9)

z/y = r6

y/x = z/y

Thus x, y, z are in G.P.

1. Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Solution:

The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. However, it can be changed to G.P. as

Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

= 8/9 [9 +99 + 999+ 9999+ … to n terms]

= 8/9 [(10-1) +(100-1) + (1000-1) + (1000-1) + … to n terms]

= 8/9 [(10 +102 + 103 + … to n terms)-(1+1+1+…n terms)]

= 8/9 [10(10^n-1)/9 – n]

= 80/81 (10n – 1) – 8/9 n

1. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 12 .

Solution:

Required sum = 2×128 + 4×32 + 8×8 +16×2 +32x12

=64[4+2+1+1/2 +1/22]

Here 4, 2, 1, 1/2, 1/22 is a G.P.

First term a = 4

Common ratio r = 1/2 Required sum = 64(31/4) = 16(31) = 496

1. Show that the products of the corresponding terms of the sequnnces form a, ar, ar2,…arn-1 and A, AR,AR2,…,ARn-1 a G.P. and find the common ratio.

Solution:

It has to be proved that the sequence: aA, arAR, ar2AR2,…arn-1ARn-1 forms a G.P.

Second term/first term = arAR/aA = rR

Third term /secondterm = ar2AR2/arAR = rR

Thus the above sequence forms a G.P. and the common ratio is rR

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